Chapter 5 of 14

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Uttar Pradesh (UPMSP)Class 11 Physics★ Unit IV · part of the 17-mark block covering Units III to VI in CBSE Class 11 Physics (70 marks total); collisions and energy conservation are key for JEE and NEET

Work, Energy and Power

Newton's laws describe motion force by force, direction by direction. This chapter throws that machinery away and asks a single scalar question instead — how much work was done? — and shows that the answer alone predicts a final speed, without ever needing to know the exact shape of the force involved. The same idea then explains why momentum survives every collision but kinetic energy only survives the gentle ones. CBSE Class 11 Physics Chapter 5.

⏱ 16 min readMedium difficulty✓ Free · NCERT-aligned

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The chapter at a glance

Every concept and exam-favourite branch — revisable in 5 minutes.

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Study Guide

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Key points, glossary and the questions most likely to show up.

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By the end of this chapter you'll be able to…

1Compute work done by constant and variable forces
2Apply the work-energy theorem (W_net = delta-K)
3Use conservation of mechanical energy for conservative systems
4Distinguish conservative and non-conservative forces
5Analyse elastic and inelastic collisions and compute power

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Why this chapter matters

Energy gives a scalar approach to mechanics that often beats force analysis. The work-energy theorem, conservation of mechanical energy, and collision laws are universal tools used far beyond mechanics, in thermodynamics and electromagnetism too.

Before you start — revise these

A 5-minute refresher here will save you 30 minutes of confusion below.

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Class 11 Physics: Laws of Motion

Force, Newton's third law, and impulse — used directly to show momentum is conserved in every collision

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Class 11 Physics: Motion in a Straight Line

The kinematic equation v^2 = u^2 + 2as is the starting point for deriving the work-energy theorem

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Class 11 Physics: Motion in a Plane

The scalar (dot) product of two vectors, used to define work as F.d

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Class 11 Mathematics: Integrals

Needed to find the area under a force-displacement graph for a variable force, and to derive the work-energy theorem in general

Work, Energy and Power

1. What this chapter covers

Textbook section	Topic
5.1-5.2	The scalar product of two vectors; work and kinetic energy linked via kinematics
5.3-5.4	Work done by a constant force; kinetic energy
5.5-5.6	Work done by a variable force, and the work-energy theorem proved for it
5.7-5.8	Potential energy; conservation of mechanical energy
5.9	Potential energy of a spring
5.10	Power
5.11	Collisions — elastic and inelastic

The chapter's own opening line is worth holding onto: work, energy and power are everyday words, but physics gives each one one precise meaning, and the aim of the whole chapter is to nail that meaning down.

2. Work needs a direction check, not just a force

The scalar product

Force and displacement are both vectors, but work is defined through their dot product:

$W = F \cdot d = F d cos θ$

where $θ$ is the angle between the force and the displacement. A dot product of two vectors always returns a scalar — work has no direction of its own, only a sign.

Reading the sign

$θ$	$cos θ$	Work	Example
$0°$	$1$	Positive — maximum	Pushing a box in the direction it moves
$90°$	$0$	Zero	Carrying a bag horizontally: force is up, displacement is forward
$180°$	$- 1$	Negative	Friction opposing motion

Zero work is easy to miss. A waiter carrying a tray at constant height does real muscular effort, but the force is vertical while the displacement is horizontal — the physics answer is exactly zero, whatever the physiology feels like.

Negative work is not a penalty term. It simply means the force point the opposite way to the motion. Gravity does negative work on a ball thrown upward and positive work on the same ball falling back down — same force, same direction, different sign only because the displacement reversed.

Work is a scalar with SI unit the joule (J) = N·m, and dimensions $[M L^{2} T^{- 2}]$ .

3. The work-energy theorem, first from kinematics

Take a constant force $F$ acting on a mass $m$ over a straight-line displacement $s$ , starting at speed $u$ and ending at speed $v$ . From Chapter 2's own kinematic equation:

$v^{2} - u^{2} = 2 a s$

Multiply both sides by $m /2$ and use $F = ma$ :

$\frac{1}{2} m v^{2} - \frac{1}{2} m u^{2} = ma s = F s = W$

The work done by the net force equals the change in kinetic energy. This is the work-energy theorem, and kinetic energy itself falls straight out of the derivation:

$K = \frac{1}{2} m v^{2}$

Kinetic energy is the energy a body has purely by virtue of its motion. Like work, it is a scalar, always $\geq 0$ , with the same unit and dimensions as work.

A genuinely useful trick the theorem enables: you can find the work done by an unknown force without ever knowing its exact form, provided you can measure the speed change it produced. A falling raindrop with air resistance of unknown shape is a textbook example — measure the impact speed, compute $Δ K$ , subtract the (known) work done by gravity, and whatever is left over is exactly the work the resistive force did.

4. Work done by a variable force

Most real forces are not constant. Split the displacement into small steps $Δ x$ over which the force is approximately constant:

$Δ W \approx F (x) Δ x$

Add up all the strips, then shrink $Δ x \to 0$ : the sum becomes the area under the force-displacement graph, and the sum becomes an integral —

$W = \int_{x_{i}}^{x_{f}} F (x) d x$

Worked, mirroring the textbook's own Example 5.5

A woman pushes a trunk along a rough platform. For the first 10 m she applies a steady 100 N; after that she tires, and her force falls off linearly from 100 N to 50 N over the next 10 m. Friction opposes her at a constant 50 N throughout the full 20 m.

Her work is the area under her own force-vs-displacement graph: a rectangle for the first 10 m plus a trapezium for the second 10 m.

$W_{F} = (100) (10) + \frac{1}{2} (100 + 50) (10) = 1000 + 750 = 1750 J$

Friction's work is a constant $- 50$ N over the full 20 m, since it opposes motion throughout:

$W_{f} = (- 50) (20) = - 1000 J$

Reading work off a graph as an area works for any shape of force curve — constant, linear, or otherwise — which is exactly why the method matters beyond this one problem.

5. The work-energy theorem, proved again for a variable force

The kinematic derivation in section 3 assumed $a$ was constant. For a genuinely variable force, go back to first principles using $d W = F d x = m \frac{d v}{d t} d x = m v d v$ (since $d x / d t = v$ ):

$W = \int_{u}^{v} m v d v = \frac{1}{2} m v^{2} - \frac{1}{2} m u^{2}$

The theorem holds exactly, whether or not the force is constant. That it survives this second, more general proof is precisely why it is such a powerful shortcut — it never needs to be re-checked case by case.

6. Potential energy: work stored, not spent

A stretched bowstring, a compressed spring, a fault line under geological strain — each has done nothing yet, but each is capable of doing work the moment it is released. That capacity, stored by virtue of position or configuration, is potential energy.

Gravitational potential energy

Raise a mass $m$ through height $h$ near Earth's surface (where $g$ can be treated as constant). The external agent does work $m g h$ against gravity, and that work is defined to be stored as potential energy:

$V (h) = m g h$

Differentiating the other way recovers the force itself:

$F = - \frac{d V}{d h} = - m g$

The minus sign is not decorative — it records that the force points opposite to the direction in which potential energy increases.

Which forces even get a potential energy?

Only conservative forces. A force is conservative when the work it does between two points depends only on those two points, never on the path taken between them. Gravity and the ideal spring force both qualify: slide a block down any smooth frictionless incline of height $h$ and it arrives at the bottom with exactly the same speed, $2 g h$ , regardless of the incline's angle.

Friction fails this test outright — a longer path against friction always costs more work — so no potential energy can be assigned to friction at all, not a small one, none.

A consequence worth memorising: the work done by a conservative force around any closed loop is exactly zero. Comets return to the same point in their orbit with the same speed every time, even though the Sun's gravity is never perpendicular to their velocity along most of the path — because gravity is conservative, and a closed loop always nets to zero work.

7. Conservation of mechanical energy

Combine the work-energy theorem ( $Δ K = F (x) Δ x$ ) with the definition of potential energy ( $- Δ V = F (x) Δ x$ ) for the same conservative force, and the force cancels out entirely:

$Δ K = - Δ V ⟹ Δ (K + V) = 0 ⟹ K + V = constant$

Define total mechanical energy $E = K + V$ . Whenever only conservative forces act, $E$ stays fixed — kinetic and potential energy trade off against each other, but their sum never changes.

Worked: a pendulum released from an angle

A bob of mass $m$ hangs from a string of length $L$ and is released from rest at angle $θ_{0}$ to the vertical. The height it falls to reach the lowest point is $h = L (1 - cos θ_{0})$ .

Since only gravity (and string tension, which does zero work — it is always perpendicular to the motion) acts, mechanical energy is conserved:

$m g h = \frac{1}{2} m v^{2} ⟹ v = 2 g L (1 - cos θ_{0})$

For $L = 1$ m and $θ_{0} = 30°$ : $h = 1 (1 - cos 30°) = 1 (1 - 0.866) = 0.134$ m, giving $v = 2 (10) (0.134) \approx 1.64$ m/s.

Tension never enters the energy equation at all — it does zero work at every instant, since it is always perpendicular to the bob's velocity. That is what makes the energy method faster here than resolving forces along the string.

8. The potential energy of a spring — and what a PE graph actually tells you

An ideal spring obeys Hooke's law, $F_{s} = - k x$ , where $x$ is displacement from the natural length. Since this force is not constant, its work is the area of a triangle on the $F$ - $x$ graph:

$W_{s} = \int_{0}^{x} (- k x^{'}) d x^{'} = - \frac{1}{2} k x^{2} ⟹ V (x) = \frac{1}{2} k x^{2}$

Plotting $V (x)$ gives an upward parabola; plotting $K (x)$ for a fixed total energy $E$ gives a downward one, and the two are exact complements — wherever $V$ rises, $K$ falls by the same amount, since $K + V = E$ stays fixed.

Why a PE graph can forbid a region outright

A particle with total energy $E$ can only exist where $V (x) \leq E$ , because $K = E - V (x)$ can never be negative. For the spring, $E = 1$ J and $k = 0.5$ N/m means the particle physically cannot go past $x = \pm 2$ m:

$\frac{1}{2} (0.5) x^{2} = 1 ⟹ x^{2} = 4 ⟹ x = \pm 2 m$

At those points $K = 0$ and the particle turns back — these are called turning points, and this is exactly how an oscillator is confined without any wall ever touching it.

9. Power: how fast the work gets done

Average power is work over time; instantaneous power is its limit as the interval shrinks to zero:

$P_{a v g} = \frac{W}{t} P = \frac{d W}{d t} = F \cdot v$

Power is a scalar with SI unit the watt (1 W = 1 J/s), dimensions $[M L^{2} T^{- 3}]$ . One horsepower is $746$ W, still used for engines and motors.

The kilowatt-hour is a unit of energy, not power — a 100 W bulb run for 10 hours uses $100 \times 10 = 1000$ Wh $= 1$ kWh $= 3.6 \times 1 0^{6}$ J. It survives on electricity bills purely because it is a convenient size for household energy use, not because it measures a rate.

Worked, mirroring the textbook's own Example 5.10

A lift of total mass 1800 kg rises at a constant 2 m/s against a constant friction force of 4000 N. The motor must supply enough force to balance both gravity and friction:

$F = m g + F_{f} = (1800) (10) + 4000 = 22000 N$ $P = F v = (22000) (2) = 44000 W \approx 59 hp$

Constant velocity is the detail that makes this solvable in one line — it means the net force is zero, so the motor's force is fixed by gravity and friction alone, with nothing left over for acceleration.

10. Collisions: momentum always, kinetic energy sometimes

Why momentum survives every collision

During any collision the two bodies exert equal and opposite impulsive forces on each other (Newton's third law), so their momentum changes are equal and opposite too — the total momentum of the system is exactly conserved, regardless of what happens to kinetic energy. This holds even though the forces during contact can vary in a complicated way moment to moment.

Where the three collision types differ

Deformation during contact can act like a compressed spring. If it fully relaxes with no energy lost, the collision is elastic — kinetic energy is the same before and after (though not necessarily during contact, when the bodies are momentarily deformed). If the bodies stick together afterward, it is perfectly inelastic, and kinetic energy loss is at its maximum for the given momentum. Everything in between — some KE lost, but the bodies separate — is simply inelastic.

Type	Momentum	Kinetic energy	Example
Elastic	conserved	conserved	idealised billiard balls
Inelastic	conserved	partly lost	a typical car crash
Perfectly inelastic	conserved	maximally lost	two bodies that stick together

Perfectly inelastic collision, one dimension

Mass $m_{1}$ moving at $v_{1 i}$ strikes a stationary $m_{2}$ , and they move off together at $v_{f}$ :

$v_{f} = \frac{m _{1} v _{1 i}}{m _{1} + m _{2}} Δ K = \frac{1}{2} \frac{m _{1} m _{2}}{m _{1} + m _{2}} v_{1 i}^{2}$

The kinetic-energy loss is always positive, as it must be — energy is lost to heat, sound and deformation, never gained.

Elastic collision, one dimension

Solving momentum and kinetic-energy conservation together for $m_{1}$ hitting stationary $m_{2}$ :

$v_{1}^{'} = \frac{m _{1} - m _{2}}{m _{1} + m _{2}} v_{1 i} v_{2}^{'} = \frac{2 m _{1}}{m _{1} + m _{2}} v_{1 i}$

Two special cases worth memorising, because they are the ones exams actually ask about:

Equal masses ( $m_{1} = m_{2}$ ): $v_{1}^{'} = 0$ , $v_{2}^{'} = v_{1 i}$ — the incoming ball stops dead and the target picks up its full velocity. Any billiards player has seen this without knowing the algebra behind it.
Very unequal masses: a heavy ball hitting a light stationary one barely slows down; a light ball hitting a massive stationary one bounces back near its own incoming speed.

11. Two traps the chapter names explicitly

KE conservation applies only before-and-after, never during contact. Even in a perfectly elastic collision, the colliding bodies are momentarily deformed and their combined kinetic energy dips during the contact interval — the "conserved" statement compares the state well before impact to the state well after it, not an instant-by-instant claim.

A conservative force does positive work → potential energy decreases, not increases. Since $Δ V = - W$ , positive work done by the force always drains potential energy, converting it to kinetic energy — the two move in opposite directions by definition, and mixing this up is one of the most common sign errors in the chapter.

Summary

Work $W = F d cos θ$ is a scalar; it is zero when force and displacement are perpendicular, and negative when they oppose.
The work-energy theorem, $W_{n e t} = Δ K$ , holds for constant and variable forces alike — proved twice in the chapter, once from kinematics and once by integration.
Potential energy $V (x)$ exists only for conservative forces, defined by $F = - d V / d x$ ; friction gets none.
Mechanical energy $E = K + V$ is conserved whenever only conservative forces act.
A spring stores $V (x) = \frac{1}{2} k x^{2}$ ; a particle can never reach a point where $V (x) > E$ , which is exactly how turning points arise.
Power is the rate of doing work, $P = d W / d t = F \cdot v$ ; the kilowatt-hour is a unit of energy, not power.
Momentum is conserved in every collision. Kinetic energy is conserved only in elastic ones, and only in the before/after sense, not instant-by-instant during contact.
Equal-mass elastic collisions exchange velocities completely — the incoming body stops, the target inherits its speed.

Key formulas & results

Everything you need to memorise, in one card. Screenshot this for revision.

Work by a constant force

W = F.d = Fd cos(theta)

Scalar; zero when force is perpendicular to displacement, negative when it opposes motion

Work by a variable force

W = integral of F(x) dx from x_i to x_f

The area under the force-displacement graph

Kinetic energy

K = (1/2)mv^2

Always >= 0; same unit and dimensions as work

Work-energy theorem

W_net = delta-K = K_f - K_i

Holds for constant AND variable forces

Gravitational potential energy

V(h) = mgh, so F = -dV/dh = -mg

Defined only near Earth's surface, where g is roughly constant

Force from potential energy

F(x) = -dV/dx

Only definable for a conservative force; friction gets no potential energy

Conservation of mechanical energy

E = K + V = constant

Holds only while every force doing work is conservative

Spring potential energy

V(x) = (1/2)kx^2, from F_s = -kx

Hooke's law; k is the spring constant, N/m

Power

P = dW/dt = F.v

SI unit watt; 1 hp = 746 W; 1 kWh = 3.6e6 J is a unit of ENERGY, not power

Perfectly inelastic collision (1D)

v_f = m1 v1i / (m1 + m2)

Bodies move off together; KE loss is maximum for the given momentum

KE loss in a perfectly inelastic collision

delta-K = (1/2) [m1 m2 / (m1+m2)] v1i^2

Always positive: lost to heat, sound and deformation

Elastic collision (1D), final velocities

v1' = [(m1-m2)/(m1+m2)] v1i ; v2' = [2m1/(m1+m2)] v1i

For m1 hitting a stationary m2

Coefficient of restitution

e = (v2' - v1')/(u1 - u2)

e = 1 elastic, e = 0 perfectly inelastic

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Common mistakes & fixes

These are the exact errors that cost students marks in board exams. Read them once, save yourself the trouble.

WATCH OUT

✗ Assuming applied force always does work

✓ Work is zero if displacement is zero or perpendicular to the force (e.g. carrying a bag horizontally).

WATCH OUT

✗ Assuming kinetic energy is conserved in all collisions

✓ Momentum is conserved in all collisions; kinetic energy is conserved only in elastic collisions.

WATCH OUT

✗ Confusing power with energy

✓ Power is the rate of energy transfer; kWh is a unit of energy, not power.

WATCH OUT

✗ Thinking potential energy cannot be negative

✓ Kinetic energy is always non-negative, but potential energy is reference-dependent and can be negative.

WATCH OUT

✗ Assuming a positive-work conservative force increases potential energy

✓ It is the opposite: delta-V = -W. When a conservative force does positive work, potential energy DECREASES by exactly that amount, converting into kinetic energy.

WATCH OUT

✗ Believing kinetic energy is conserved at every instant during an elastic collision

✓ It is conserved only comparing well-before to well-after. During contact the bodies are momentarily deformed, and combined KE dips below its before/after value — the theorem never claims otherwise.

WATCH OUT

✗ Forgetting that string/rod tension does zero work in a pendulum or circular problem

✓ Tension is always perpendicular to the instantaneous velocity, so it contributes nothing to the work-energy or energy-conservation equation — only gravity (or another applied force) needs to be tracked.

WATCH OUT

✗ Assuming a particle can be found anywhere on a potential energy graph

✓ A particle of total energy E can only exist where V(x) <= E, since K = E - V(x) can never be negative. Points where V(x) = E are turning points; regions where V(x) > E are physically forbidden.

WATCH OUT

✗ Using W = F d cos(theta) for a force that changes during the motion

✓ That formula assumes F is constant. For a variable force, work is the area under the F-x graph, or the integral of F(x) dx — treat each small step as approximately constant and sum (or integrate).

WATCH OUT

✗ Thinking the spring force's sign and the displacement's sign always tell you the work is positive

✓ The spring force Fs = -kx and the displacement x always point opposite once you compress or stretch and let go, so the spring's own work is negative on the way out and positive only when an external force does the stretching against it.

WATCH OUT

✗ Assuming friction can be assigned a potential energy like gravity or a spring

✓ Friction is non-conservative — work done against it depends on the path taken and is nonzero over a closed loop, so no potential energy function can be defined for it, ever.

NCERT exercises (with solutions)

Every NCERT exercise from this chapter — what it covers and how many questions to expect.

Ex s

Exercises

Sign of work, potential energy diagrams, work-energy theorem, power, and elastic/inelastic collisions (Q5.1-5.23)

Questions

Practice problems

Work through this chapter's problems as a readiness check — reveal each solution, mark yourself honestly, and get your gap report at the end.

Readiness check

Are you exam-ready for Work, Energy and Power?

10 problems from this chapter. Try each one, reveal the worked solution, mark yourself honestly — get your gap report at the end.

10 questions~7 min worth ~17 marks in Uttar Pradesh (UPMSP) exams

5-minute revision

The whole chapter, distilled. Read this the night before the exam.

•Work W = Fd cos(theta); scalar; zero when force is perpendicular to displacement.
•Work-energy theorem: net work = change in kinetic energy (for any force).
•K = (1/2)mv^2; gravitational PE = mgh; spring PE = (1/2)kx^2.
•Mechanical energy is conserved when only conservative forces act.
•Conservative forces do zero work over a closed path; friction is non-conservative.
•Power P = W/t = F.v; SI unit watt.
•Momentum is conserved in all collisions; KE only in elastic ones.

Uttar Pradesh (UPMSP) marks blueprint

Where the marks come from in this chapter — so you can plan your prep.

Typical chapter weightage: Unit IV sits inside the 17-mark block covering Units III to VI (CBSE Class 11 Physics, 70 marks)

Question type	Marks each	Typical count	What it tests
Work and the work-energy theorem	2-3	1	Sign of work in a given situation; work by a variable force as an area/integral; the work-energy theorem applied to find a final speed
Potential energy conservation of mechanical energy and spring energy	3-5	1-2	Gravitational and spring potential energy; conservation of mechanical energy in pendulum/incline problems; reading forbidden regions and turning points off a potential-energy graph
Power and collisions	3-5	1	Average and instantaneous power; elastic and inelastic collisions in one dimension; coefficient of restitution

Prep strategy

Use the work-energy theorem to shortcut force problems
Apply energy conservation to pendulum and incline problems
Learn elastic-collision velocity formulae and special cases
Distinguish power from energy and watch units

Where this shows up in the real world

This chapter isn't just an exam topic — it lives in the world around you.

Roller coasters

Energy conversion between kinetic and gravitational potential energy designs thrilling, safe rides.

Vehicle crash testing

Collision analysis predicts energy absorbed in crashes and informs safety design.

Power ratings

Power calculations rate engines, motors, and appliances and govern electricity billing in kWh.

Exam strategy

Battle-tested tips from teachers and toppers for this chapter.

Choose energy methods to avoid messy force resolution

Set a clear reference level for potential energy

State whether a collision is elastic before assuming KE conservation

Keep power and energy units distinct

Going beyond the textbook

For olympiad aspirants and curious learners — topics that build on this chapter.

STRETCH

Analyse two-dimensional collisions using momentum components and restitution.

STRETCH

Derive escape conditions and orbital energy using energy conservation.

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JEE Main & Advanced practice

Competitive-level problems on this chapter, above the board pattern. Try each one on paper before opening the solution.

JEE MainWork-energy theorem with friction on an inclineFormula application

A 2 kg block slides down a rough incline of length 5 m, inclined at $30°$ to the horizontal, with coefficient of kinetic friction $μ_{k} = 0.2$ . Find the speed of the block at the bottom, using the work-energy theorem. (Take g = 10 m/s^2.)

Stuck? Show the approach

Find the work done by gravity along the incline and the work done against friction separately, add them as scalars to get the net work, then set that equal to the gain in kinetic energy.

Show the full solution

$W_{g r a v i t y} = m g s sin θ = (2) (10) (5) (0.5) = 50 J$ $W_{f r i c t i o n} = - μ_{k} m g s cos θ = - (0.2) (2) (10) (5) (cos 30°) \approx - 17.3 J$ $W_{n e t} = 50 - 17.3 = 32.7 J = Δ K = \frac{1}{2} (2) v^{2} ⟹ v \approx 5.72 m/s$

Answer: approximately 5.72 m/s

The trap

Forgetting that the normal force used to compute friction is mg cos(theta), not mg, is the most common slip — on an incline the friction force scales with the perpendicular component of weight, not the full weight.

JEE MainPower against a speed-dependent resistive forceFormula application

A car of mass 1000 kg moves at a constant velocity of 20 m/s against a resistive force $F = k v^{2}$ , where $k = 0.5$ N·s²/m². Find the power delivered by the engine.

Stuck? Show the approach

At constant velocity the net force is zero, so the engine's driving force exactly equals the resistive force at that speed. Compute that force first, then use P = Fv.

Show the full solution

$F = k v^{2} = (0.5) (20)^{2} = 200 N$ $P = F v = (200) (20) = 4000 W = 4 kW$

Answer: 4000 W (4 kW)

The trap

Using the car's mass or trying to apply F = ma is a red herring here — at constant velocity the acceleration is zero, so the mass never enters the calculation at all.

JEE MainCoefficient of restitution from bounce heightsFormula application

A ball is dropped from a height of 5 m onto a hard floor and bounces back to a height of 3.2 m. Find the coefficient of restitution.

Stuck? Show the approach

Convert both heights to speeds using energy conservation (v = sqrt(2gh)), then take the ratio of the rebound speed to the impact speed — that ratio is the coefficient of restitution for a collision with a stationary, immovable floor.

Show the full solution

$v_{im p a c t} = 2 g h_{1}, v_{r e b o u n d} = 2 g h_{2} ⟹ e = \frac{v _{r e b o u n d}}{v _{im p a c t}} = \frac{h _{2}}{h _{1}} = \frac{3.2}{5} = 0.64 = 0.8$

Answer: e = 0.8

The trap

Taking e as the ratio of heights directly (h2/h1 = 0.64) instead of the square root of that ratio is the standard error — restitution compares speeds, and speed goes as the square root of height, not height itself.

JEE AdvancedMotion under a source of constant powerDifferential-equation derivation

A body of mass m, starting from rest, moves under the action of a source delivering constant power P. Show that its velocity varies as $v \propto t^{1/2}$ and the distance it has covered varies as $s \propto t^{3/2}$ .

Stuck? Show the approach

Power is not force, so none of the standard constant-acceleration kinematic equations apply directly. Start from P = Fv = m(dv/dt)v, which is a differential equation in v and t, and separate variables.

Show the full solution

$P = m v \frac{d v}{d t} ⟹ v d v = \frac{P}{m} d t ⟹ \int_{0}^{v} v d v = \frac{P}{m} \int_{0}^{t} d t ⟹ \frac{v ^{2}}{2} = \frac{P t}{m} ⟹ v = \frac{2 P t}{m}$

So $v \propto t^{1/2}$ . Integrating again for distance:

$s = \int_{0}^{t} \frac{2 P}{m} t^{1/2} d t = \frac{2 P}{m} \cdot \frac{2}{3} t^{3/2} ⟹ s \propto t^{3/2}$

Answer: v is proportional to the square root of t; s is proportional to t raised to the power 3/2

The trap

Trying to use v = u + at is the reflex mistake — constant power means the force, and hence the acceleration, is continuously decreasing as speed rises (since F = P/v), so none of the three standard kinematic equations are valid here.

JEE AdvancedPerpendicular paths after an equal-mass glancing elastic collisionVector derivation

A ball of mass m moving with velocity $v_{0}$ strikes an identical stationary ball of mass m in a glancing (two-dimensional) elastic collision. Prove that, unless the collision is exactly head-on, the two balls move off at right angles to each other after the collision.

Stuck? Show the approach

Write momentum conservation as a single vector equation and kinetic energy conservation as a scalar equation, then compare the squared magnitude of the vector momentum equation against the scalar energy equation — the cross term that survives that comparison is exactly the dot product of the two outgoing velocities.

Show the full solution

Momentum: $m v_{0} = m v_{1} + m v_{2} ⟹ v_{0} = v_{1} + v_{2}$ .

Squaring (dotting with itself): $v_{0}^{2} = v_{1}^{2} + v_{2}^{2} + 2 v_{1} \cdot v_{2}$ .

Kinetic energy conservation (equal masses): $v_{0}^{2} = v_{1}^{2} + v_{2}^{2}$ .

Comparing the two: $2 v_{1} \cdot v_{2} = 0 ⟹ v_{1} \cdot v_{2} = 0$ , so $v_{1}$ and $v_{2}$ are perpendicular, provided neither is zero (which excludes only the exactly head-on case, where the target simply takes over $v_{0}$ entirely).

Answer: v1 . v2 = 0, so the two balls' paths are perpendicular after a glancing equal-mass elastic collision

The trap

This result depends critically on the masses being equal AND the target being initially at rest — change either condition and the neat perpendicularity disappears, so it cannot be assumed as a general collision fact.

Where else this chapter is tested

CBSE board isn't the only one — other exams test this chapter too.

CBSE Class 11 Physics examHigh

JEE Main and Advanced (Work-Energy)High

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Questions students ask

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In both kinds, total linear momentum is conserved. In an elastic collision, total kinetic energy is also conserved (e.g. ideal billiard balls), with coefficient of restitution e = 1. In an inelastic collision some kinetic energy is converted into heat, sound, or deformation, so KE is not conserved; in a perfectly inelastic collision the bodies stick together (e = 0) and the KE loss is maximum.

A conservative force, such as gravity or a spring force, depends only on position and can be associated with a potential energy. The work it does going from A to B is the negative of the change in potential energy, independent of path. Returning to the start brings the potential energy back to its original value, so the net work over the closed loop is exactly zero -- which is the defining property of a conservative force.

Yes -- in fact that is exactly when it saves the most work. Add up all the forces acting (gravity, friction, an applied push, a normal force) as vectors first to get the net force, or equivalently compute the work done by each force separately and add those numbers, since work is a scalar and adds directly. Either route gives the same total change in kinetic energy. What the theorem will not tell you is how that kinetic energy change was split between different causes unless you compute each force's work individually, as the raindrop example in this chapter does.

Its kinetic energy is exactly zero at that point, since K = E - V(x), so the particle is momentarily at rest -- but not permanently stuck. Unless that point happens to sit at a force-free equilibrium, the force F = -dV/dx is still nonzero there, so the particle immediately accelerates back the way it came. This is exactly a turning point: the particle reaches it, stops for an instant, and reverses, the same way a ball thrown upward stops for an instant at its highest point before falling back.

Solving the general elastic-collision formulas with m1 = m2 makes the (m1-m2) term in the expression for v1' vanish entirely, leaving v1' = 0, while the 2m1/(m1+m2) term in v2' becomes exactly 1, leaving v2' = v1i. The incoming ball transfers all of its momentum and all of its kinetic energy to the target and stops dead, while the target inherits the incoming ball's full velocity. This is the everyday experience of striking a stationary billiard ball of the same mass head-on.

It is a more precise, continuous version of the same idea. The coefficient of restitution e is the ratio of the relative speed of separation to the relative speed of approach. For a perfectly elastic collision e = 1 (the bodies separate exactly as fast as they approached); for a perfectly inelastic collision e = 0 (they do not separate at all, since they move off together); every value strictly between 0 and 1 describes an ordinary inelastic collision, with more kinetic energy lost as e drops closer to 0.

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